Lorenz Curve for a Normal Distribution The Lorenz Curve (https://en.wikipedia.org/wiki/Lorenz_curve) is a visual representation of the equality of income distribution across a population. It plots the cumulative fraction of total income versus the bottom fraction of population that earns it, so both the domain (fraction of population) and range (fraction of total income) extend from 0 to 1. I'd like to generate the Lorenz curve for various income distributions (just for brain-limbering), starting with normal.
From the linked article, for a distribution with PDF f(x), CDF F(x)
, and mean $\mu $, the associated Lorenz Curve is 
$$L\left( {F\left( x \right)} \right) = {1 \over \mu }\int\limits_{ - \infty }^x {t\;f\left( t \right)dt} $$
For the normal distribution, $f\left( x \right) = \phi \left( x \right)$, $F\left( x \right) = \Phi \left( x \right)$, and therefore
$$L\left( {\Phi \left( x \right)} \right) = {1 \over \mu }\int\limits_{ - \infty }^x {t\;\phi \left( t \right)dt} $$
Refreshing an ancient memory of the Product Rule for integration,
$$\int {u\;dv}  = uv - \int {v\;du} $$
So let $u = t$ and ${dv = \phi \left( t \right)dt}$; then $du=dt$ and ${v = \Phi \left( t \right)}$. Then
$$L\left( {\Phi \left( x \right)} \right) = {1 \over \mu }\left( {\left. {t\;\Phi \left( t \right)} \right|_{ - \infty }^x - \int\limits_{ - \infty }^x {\Phi \left( t \right)dt} } \right) = {1 \over \mu }\left( {t\;\Phi \left( x \right) - \int\limits_{ - \infty }^x {\Phi \left( t \right)dt} } \right)$$
Assuming that's correct so far, I've run out of steam. Is there a common function to evaluate that definite integral on the right?
Thanks for reading.
 A: actually for normal distribution you can do the integration directly. this is just another way.
$\alpha=\frac{a-\mu}{\sigma}=-\infty, \beta=\frac{b-\mu}{\sigma}=\frac{x-\mu}{\sigma}$
$\mu'\equiv \frac{\int_{-\infty}^{x}tf(t)dt}{\int_{-\infty}^{x}f(t)dt}=\mu+\frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)}\sigma=\mu-\frac{\phi(\beta)}{\Phi(\beta)}\sigma$
$L(F(x))=\frac{1}{\mu}\int_{-\infty}^{x}tf(t)dt=\frac{\mu'\Phi(\beta)}{\mu}=\Phi(\beta)-\frac{\phi(\beta)}{\mu}\sigma$
The direct way:
$L(F(x))=\frac{1}{\mu}\int_{-\infty}^{x}tf(t)dt=\frac{1}{\mu}(\mu\Phi(\beta)+\int_{-\infty}^{x}(t-\mu)f(t)dt)$
$=\Phi(\beta)+\frac{1}{\mu\sqrt{2\pi\sigma^2}}\int_{-\infty}^{x}(t-\mu)\cdot e^{-\frac{(t-\mu)^2}{2\sigma^2}}\cdot dt$
$=\Phi(\beta)+\frac{\sigma}{\mu\sqrt{2\pi}}\int_{-\infty}^{x}\frac{t-\mu}{\sigma}\cdot e^{-\frac{(t-\mu)^2}{2\sigma^2}}\cdot d{\frac{t-\mu}{\sigma}}$
$=\Phi(\beta)-\frac{\sigma}{\mu\sqrt{2\pi}}\int_{-\infty}^{\beta}d{e^{-\frac{t.'^2}{2}}}$
$=\Phi(\beta)-\frac{\sigma}{\mu}\frac{1}{\sqrt{2\pi}}e^{-\frac{\beta^2}{2}}$
$=\Phi(\beta)-\frac{\sigma}{\mu}\phi(\beta)$
